f(x) = \frac{1}{\frac{1}{4} \sin^2 2x} + 1 = \frac{4}{\sin^2 2x} + 1
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- f(x) = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} + 1 = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} + 1 = \frac{1}{\sin^2 x \cos^2 x} + 1
- Using the identity $ \sin(2x) = 2\sin x \cos x $, we have:
- \sin x \cos x = \frac{1}{2} \sin 2x \Rightarrow \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x
- Since $ 0 < x < \frac{\pi}{2} $, then $ 0 < 2x < \pi $, and $ \sin 2x > 0 $. The minimum of $ f(x) $ occurs when $ \sin^2 2x $ is maximized, i.e., when $ \sin^2 2x = 1 $. This happens when $ 2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4} $.
- f\left(\frac{\pi}{4}\right) = \frac{4}{1} + 1 = 5
- Thus, the minimum value is $ \boxed{5} $.
